2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

Given:
\(u_2 = a(2)^2 + b(2) - 12 = 18\)
\(4a + 2b - 12 = 18 \implies 4a + 2b = 30 \implies 2a + b = 15\) (Equation 1)

And:
\(u_5 = a(5)^2 + b(5) - 12 = 138\)
\(25a + 5b - 12 = 138 \implies 25a + 5b = 150 \implies 5a + b = 30\) (Equation 2)

Subtracting Equation 1 from Equation 2:
\((5a + b) - (2a + b) = 30 - 15\)
\(3a = 15 \implies a = 5\)

Substituting \(a = 5\) into Equation 1:
\(2(5) + b = 15 \implies 10 + b = 15 \implies b = 5\)

So, the \(n\)-th term is:
\(u_n = 5n^2 + 5n - 12\)

To find the 10th term, \(u_{10}\):
\(u_{10} = 5(10)^2 + 5(10) - 12 = 5(100) + 50 - 12 = 500 + 50 - 12 = 538\)

M1: For setting up two simultaneous equations in terms of \(a\) and \(b\), e.g., \(4a + 2b = 30\) and \(25a + 5b = 150\) (or equivalent).
A1: For correctly solving to find \(a = 5\) and \(b = 5\).
M1: For substituting \(n = 10\) and their values of \(a\) and \(b\) into the formula for \(u_n\).
A1.7: For the correct final answer of 538.

Let the \(n\)-th term of the geometric sequence be \(u_n = a r^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.
We are given:
\(u_3 = a r^2 = 18\)
\(u_5 = a r^4 = 40.5\)

Dividing the equation for \(u_5\) by the equation for \(u_3\):
\(\frac{a r^4}{a r^2} = \frac{40.5}{18}
\)r^2 = 2.25\)

Since the common ratio \(r\) is positive, we take the positive square root:
\(r = \sqrt{2.25} = 1.5\)

Now, substitute \(r = 1.5\) back to find \(a\):
\(a(1.5)^2 = 18 \implies 2.25 a = 18 \implies a = 8\)

The sum of the first \(n\) terms of a geometric sequence is given by:
\(S_n = \frac{a(r^n - 1)}{r - 1}

For \)n = 6\):
\(S_6 = \frac{8(1.5^6 - 1)}{1.5 - 1} = \frac{8(11.390625 - 1)}{0.5} = 16 \times 10.390625 = 166.25\)

M1: For establishing \(r^2 = \frac{40.5}{18} = 2.25\) or equivalent.
A1: For finding \(r = 1.5\) and the first term \(a = 8\).
M1: For using the sum of a geometric series formula with \(n = 6\).
A1.7: For the correct answer 166.25.

To find the vertical asymptotes, we determine the values of \(x\) for which the denominator is zero (and the numerator is non-zero).
Set the denominator to zero:
\(2x^2 - 5x - 3 = 0\)

Solving this quadratic equation:
\((2x + 1)(x - 3) = 0\)
This gives \(x = -0.5\) and \(x = 3\).

Checking the numerator, \(3x^2 - 12\), at these values:
For \(x = -0.5\), \(3(-0.5)^2 - 12 = 3(0.25) - 12 = -11.25 \neq 0\).
For \(x = 3\), \(3(3)^2 - 12 = 27 - 12 = 15 \neq 0\).
Thus, both are vertical asymptotes. Since \(a < b\), we have:
\(a = -0.5\) and \(b = 3\).

To find the horizontal asymptote, we examine the limit as \(x \to \pm\infty\):
\(y = \lim_{x \to \pm\infty} \frac{3x^2 - 12}{2x^2 - 5x - 3} = \frac{3}{2} = 1.5\)
So, \(c = 1.5\).

We need to find \(a + b + c\):
\(a + b + c = -0.5 + 3 + 1.5 = 4\)

M1: For setting denominator to 0 and solving to find \(x = -0.5\) and \(x = 3\) (or equivalent).
A1: For identifying \(a = -0.5\) and \(b = 3\).
M1: For finding the horizontal asymptote \(y = 1.5\) (or \(c = 1.5\)) by considering the coefficients of \(x^2\).
A1.7: For correctly calculating \(a + b + c = 4\).

The vertical asymptote occurs when the denominator is equal to zero, provided the numerator is non-zero.
\(q - 3x = 0 \implies x = \frac{q}{3}

Since the vertical asymptote is given as \)x = 4\):
\(\frac{q}{3} = 4 \implies q = 12\)

The horizontal asymptote is found by looking at the behavior of \(f(x)\) as \(x \to \pm\infty\):
\(y = \lim_{x \to \pm\infty} \frac{px + 7}{q - 3x} = \frac{p}{-3} = -\frac{p}{3}

Since the horizontal asymptote is given as \)y = -2\):
\(-\frac{p}{3} = -2 \implies p = 6\)

Now, we calculate \(p + q\):
\(p + q = 6 + 12 = 18\)

M1: For using the vertical asymptote condition \(q - 3x = 0\) with \(x = 4\).
A1: For finding \(q = 12\).
M1: For using the horizontal asymptote condition \(-\frac{p}{3} = -2\).
A1.7: For finding \(p = 6\) and computing \(p + q = 18\).

We use the area formula for a triangle:
\(\text{Area} = \frac{1}{2} a c \sin B\)

Here, \(a = 12.4\), \(c = 8.5\), and \(\text{Area} = 42.5\).
\(42.5 = \frac{1}{2} \times 12.4 \times 8.5 \times \sin(\angle ABC)\)
\(42.5 = 52.7 \times \sin(\angle ABC)\)
\(\sin(\angle ABC) = \frac{42.5}{52.7} \approx 0.80645\)

Since angle \(ABC\) is obtuse:
\(\angle ABC = 180^\circ - \arcsin(0.80645) \approx 180^\circ - 53.748^\circ = 126.252^\circ\)

Now we use the Cosine Rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC)\)
\(AC^2 = 8.5^2 + 12.4^2 - 2(8.5)(12.4)\cos(126.252^\circ)\)
\(AC^2 = 72.25 + 153.76 - 210.8 \times (-0.59132)\)
\(AC^2 = 226.01 + 124.65 = 350.66\)
\(AC = \sqrt{350.66} \approx 18.726\text{ cm}

Rounding to 3 significant figures, we get \)18.7\text{ cm}\).

M1: For using the area formula \(\frac{1}{2} \times 8.5 \times 12.4 \times \sin(\angle ABC) = 42.5\).
A1: For finding the obtuse angle \(\angle ABC \approx 126.252^\circ\).
M1: For applying the Cosine Rule to find \(AC^2\).
A1.7: For the correct answer 18.7 (accept 18.7 to 18.8).

First, we find the angle \(\angle PQR\) inside the triangle \(PQR\).
The bearing of \(Q\) from \(P\) is \(065^\circ\).
Therefore, the bearing of \(P\) from \(Q\) is:
\(065^\circ + 180^\circ = 245^\circ\)

The bearing of \(R\) from \(Q\) is \(140^\circ\).
The angle \(\angle PQR\) is the difference between these two bearings:
\(\angle PQR = 245^\circ - 140^\circ = 105^\circ\)

Now, we use the Cosine Rule in triangle \(PQR\) to find the distance \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)\)
\(PR^2 = 15^2 + 22^2 - 2(15)(22)\cos(105^\circ)\)
\(PR^2 = 225 + 484 - 660\cos(105^\circ)\)

Using a calculator:
\(\cos(105^\circ) \approx -0.2588\)
\(PR^2 = 709 - 660(-0.2588) \approx 709 + 170.82 = 879.82\)
\(PR = \sqrt{879.82} \approx 29.66\text{ km}

Rounding to 3 significant figures gives \)29.7\text{ km}\).

M1: For finding the bearing of \(P\) from \(Q\) as \(245^\circ\) or using parallel lines to find the interior angle at \(Q\) as \(180^\circ - 65^\circ = 115^\circ\).
A1: For finding the correct angle \(\angle PQR = 105^\circ\).
M1: For applying the Cosine Rule using their angle \(\angle PQR\).
A1.7: For the correct distance 29.7 (accept 29.6 to 29.7).

The formula is \(V = A \times k^t\).
Since the initial value (\(t = 0\)) is $12,000:
\(12000 = A \times k^0 \implies A = 12000\)

After 5 years (\(t = 5\)), the value is $17,631.94:
\(17631.94 = 12000 \times k^5\)
\(k^5 = \frac{17631.94}{12000} \approx 1.469328\)
\(k = (1.469328)^{1/5} \approx 1.08\)

So, the value grows at an annual rate of 8%, and the formula is:
\(V = 12000 \times (1.08)^t\)

To find the value after 12 years (\(t = 12\)):
\(V = 12000 \times (1.08)^{12}\)
\(V \approx 12000 \times 2.51817 = 30218.04\)

Rounding to the nearest dollar, we get $30,218.

M1: For using \(A = 12000\) and setting up the equation \(12000 \times k^5 = 17631.94\).
A1: For solving to find \(k = 1.08\) (or equivalent).
M1: For calculating the value when \(t = 12\) using their value of \(k\).
A1.7: For the correct answer 30218 (accept 30218 or 30220 with 3 s.f.).

By substituting the coordinates of the three points into the function \(f(x) = ax^2 + bx + c\), we get a system of three linear equations:
1) For \((-1, -2)\):
\(a(-1)^2 + b(-1) + c = -2 \implies a - b + c = -2\)

2) For \((2, 19)\):
\(a(2)^2 + b(2) + c = 19 \implies 4a + 2b + c = 19\)

3) For \((4, 53)\):
\(a(4)^2 + b(4) + c = 53 \implies 16a + 4b + c = 53\)

Subtract Equation (1) from Equation (2):
\((4a + 2b + c) - (a - b + c) = 19 - (-2)\)
\(3a + 3b = 21 \implies a + b = 7\) (Equation 4)

Subtract Equation (2) from Equation (3):
\((16a + 4b + c) - (4a + 2b + c) = 53 - 19\)
\(12a + 2b = 34 \implies 6a + b = 17\) (Equation 5)

Now, subtract Equation (4) from Equation (5):
\((6a + b) - (a + b) = 17 - 7\)
\(5a = 10 \implies a = 2\)

Substitute \(a = 2\) into Equation (4):
\(2 + b = 7 \implies b = 5\)

Substitute \(a = 2\) and \(b = 5\) into Equation (1):
\(2 - 5 + c = -2 \implies -3 + c = -2 \implies c = 1\)

Thus, the quadratic function is:
\(f(x) = 2x^2 + 5x + 1\)

To find \(f(5)\):
\(f(5) = 2(5)^2 + 5(5) + 1 = 2(25) + 25 + 1 = 50 + 25 + 1 = 76\)

M1: For writing down at least two correct equations in terms of \(a\), \(b\), and \(c\).
A1: For eliminating \(c\) to find two simplified equations in \(a\) and \(b\).
M1: For solving the simultaneous equations to obtain the values \(a = 2\), \(b = 5\), and \(c = 1\).
A1.7: For correctly evaluating \(f(5) = 76\).

We analyze the numerators and denominators of the sequence separately.

1. **Numerators:**
The sequence of numerators is: \(3, 8, 15, 24, 35, \dots\)
Let us find the first and second differences:
- First differences: \(5, 7, 9, 11, \dots\)
- Second differences: \(2, 2, 2, \dots\)
Since the second difference is constant and equal to \(2\), the general term is of the form \(u_n = an^2 + bn + c\) with \(a = \frac{2}{2} = 1\).
Comparing \(n^2\) with the numerator sequence:
- For \(n = 1\): \(1^2 + bn + c = 3 \implies b + c = 2\)
- For \(n = 2\): \(2^2 + 2b + c = 8 \implies 2b + c = 4\)
Subtracting the two equations gives \(b = 2\), which leads to \(c = 0\).
Thus, the \(n\)-th term of the numerator is \(n^2 + 2n\).
For \(n = 15\), the numerator is:
\[15^2 + 2(15) = 225 + 30 = 255\]

2. **Denominators:**
The sequence of denominators is: \(2, 5, 10, 17, 26, \dots\)
Comparing this sequence with perfect squares:
- \(1^2 + 1 = 2\)
- \(2^2 + 1 = 5\)
- \(3^2 + 1 = 10\)
Thus, the \(n\)-th term of the denominator is \(n^2 + 1\).
For \(n = 15\), the denominator is:
\[15^2 + 1 = 225 + 1 = 226\]

3. **15th Term:**
Combining both parts, the 15th term is:
\[\frac{255}{226}\]
Since the only prime factors of \(255\) are \(3, 5, 17\) and the prime factors of \(226\) are \(2, 113\), this fraction is already in its simplest form.

M1 for finding the formula for the numerator sequence is \(n^2 + 2n\) (or calculating the 15th numerator as 255)
M1 for finding the formula for the denominator sequence is \(n^2 + 1\) (or calculating the 15th denominator as 226)
M1 for writing the term as a fraction of the two calculated values
A1.7 for the final simplified fraction \(\frac{255}{226}\)

Let \(a = 6\) be the first term of both sequences.

1. **Arithmetic Sequence Terms:**
- The \(n\)-th term of the arithmetic sequence is given by \(u_n = a + (n-1)d\).
- \(u_3 = 6 + 2d\)
- \(u_{10} = 6 + 9d\)

2. **Geometric Sequence Terms:**
- The \(n\)-th term of the geometric sequence is given by \(g_n = a r^{n-1}\).
- \(g_2 = 6r\)
- \(g_3 = 6r^2\)

3. **Set up Equations:**
- From \(u_3 = g_2\):
\[6 + 2d = 6r \implies 2d = 6r - 6 \implies d = 3r - 3\]
- From \(u_{10} = g_3\):
\[6 + 9d = 6r^2\]

4. **Solve for \(r\):**
Substitute \(d = 3r - 3\) into the second equation:
\[6 + 9(3r - 3) = 6r^2\]
\[6 + 27r - 27 = 6r^2\]
\[6r^2 - 27r + 21 = 0\]
Divide the entire equation by 3:
\[2r^2 - 9r + 7 = 0\]
Factor the quadratic equation:
\[(2r - 7)(r - 1) = 0\]
Since we are given that \(r \neq 1\), we have:
\[2r - 7 = 0 \implies r = \frac{7}{2} = 3.5\]

M1 for writing an equation linking \(d\) and \(r\) from \(u_3 = g_2\): e.g., \(6 + 2d = 6r\)
M1 for writing an equation linking \(d\) and \(r\) from \(u_{10} = g_3\): e.g., \(6 + 9d = 6r^2\)
M1 for substituting to eliminate \(d\) and obtaining a quadratic in \(r\): e.g., \(2r^2 - 9r + 7 = 0\)
A1.7 for finding \(r = 3.5\) (or \(\frac{7}{2}\)) and rejecting \(r = 1\)

1. **Form a system of equations:**
Using the recurrence relation for the first few terms:
- For \(n = 1\):
\[x_2 = a x_1 + b \implies 14 = 4a + b \quad \text{--- (Equation 1)}\]
- For \(n = 2\):
\[x_3 = a x_2 + b \implies 44 = 14a + b \quad \text{--- (Equation 2)}\]

2. **Solve for \(a\) and \(b\):**
Subtract Equation 1 from Equation 2:
\[(14a + b) - (4a + b) = 44 - 14\]
\[10a = 30 \implies a = 3\]
Substitute \(a = 3\) back into Equation 1:
\[4(3) + b = 14 \implies 12 + b = 14 \implies b = 2\]

3. **Find subsequent terms:**
The recurrence relation is \(x_{n+1} = 3x_n + 2\).
- Calculate \(x_4\):
\[x_4 = 3x_3 + 2 = 3(44) + 2 = 132 + 2 = 134\]
- Calculate \(x_5\):
\[x_5 = 3x_4 + 2 = 3(134) + 2 = 402 + 2 = 404\]

M1 for setting up the simultaneous equations \(4a + b = 14\) and \(14a + b = 44\)
M1 for solving to find \(a = 3\) and \(b = 2\)
M1 for calculating \(x_4 = 134\)
A1.7 for obtaining the final answer of 404

1. **Find the vertical asymptotes:**
Vertical asymptotes occur where the denominator is zero and the numerator is non-zero.
Set the denominator to zero:
\[2x^2 - 5x - 3 = 0\]
Factorize the quadratic expression:
\[(2x + 1)(x - 3) = 0\]
This gives \(x = -0.5\) and \(x = 3\).
Check the numerator at these values:
\[3x^2 - 12 = 3(x-2)(x+2)\]
Since the roots of the numerator are \(x = 2\) and \(x = -2\), the numerator is non-zero at \(x = -0.5\) and \(x = 3\).
Therefore, the vertical asymptotes are \(x = -0.5\) and \(x = 3\).
Since \(a < b\), we have:
\[a = -0.5, \quad b = 3\]

2. **Find the horizontal asymptote:**
To find the horizontal asymptote, we look at the limit of \(f(x)\) as \(x \to \pm\infty\):
\[y = \lim_{x \to \pm\infty} \frac{3x^2 - 12}{2x^2 - 5x - 3} = \frac{3}{2} = 1.5\]
Thus, \(c = 1.5\).

3. **Calculate \(a + b + c\):**
\[a + b + c = -0.5 + 3 + 1.5 = 4\]

M1 for setting the denominator to 0 and attempting to solve: \(2x^2 - 5x - 3 = 0\)
M1 for finding the vertical asymptotes \(x = -0.5\) and \(x = 3\)
M1 for finding the horizontal asymptote \(y = 1.5\) (or \(y = \frac{3}{2}\))
A1.7 for summing the values to obtain \(a + b + c = 4\)

1. **Determine the vertical asymptote:**
The vertical asymptote is found by setting the denominator of the fraction to 0:
\[2x - 8 = 0 \implies x = 4\]
This matches the x-coordinate of the intersection point \((4, 11)\).

2. **Determine the horizontal asymptote:**
As \(x \to \pm\infty\), the term \(\frac{bx - 1}{2x - 8}\) approaches \(\frac{b}{2}\).
Therefore, the horizontal asymptote is:
\[y = a + \frac{b}{2}\]

3. **Use the intersection point:**
Since the horizontal asymptote is \(y = 11\), we have:
\[a + \frac{b}{2} = 11\]

4. **Substitute \(b = 2a\) and solve:**
\[a + \frac{2a}{2} = 11\]
\[a + a = 11\]
\[2a = 11 \implies a = 5.5\]

M1 for identifying that the horizontal asymptote of the curve is \(y = a + \frac{b}{2}\)
M1 for setting up the equation \(a + \frac{b}{2} = 11\) using the intersection point
M1 for substituting \(b = 2a\) into the horizontal asymptote equation
A1.7 for correctly solving to get \(a = 5.5\) (or \(\frac{11}{2}\))

1. **Find \(\sin B\) using the area formula:**
\[\text{Area} = \frac{1}{2} \times AB \times BC \times \sin B\]
\[28 = \frac{1}{2} \times 7 \times 10 \times \sin B\]
\[28 = 35 \sin B \implies \sin B = \frac{28}{35} = 0.8\]

2. **Find \(\cos B\) for the obtuse angle \(B\):**
Since angle \(B\) is obtuse (\(90^\circ < B < 180^\circ\)), its cosine must be negative.
Using the identity \(\sin^2 B + \cos^2 B = 1\):
\[\cos^2 B = 1 - 0.8^2 = 1 - 0.64 = 0.36\]
\[\cos B = -\sqrt{0.36} = -0.6\]
*(Alternatively, using a calculator: \(B = 180^\circ - \arcsin(0.8) \approx 126.87^\circ\), and \(\cos(126.87^\circ) \approx -0.6\))*.

3. **Calculate \(AC\) using the Cosine Rule:**
\[AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos B\]
\[AC^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times (-0.6)\]
\[AC^2 = 49 + 100 - 140 \times (-0.6)\]
\[AC^2 = 149 + 84 = 233\]
\[AC = \sqrt{233} \approx 15.264\text{ cm}\]

4. **Round to 3 significant figures:**
\[AC \approx 15.3\text{ cm}\]

M1 for substituting values into the area formula: \(28 = 0.5 \times 7 \times 10 \times \sin B\)
M1 for determining \(\sin B = 0.8\) and identifying \(\cos B = -0.6\) or finding angle \(B \approx 126.9^\circ\)
M1 for substituting their values into the Cosine Rule: \(AC^2 = 7^2 + 10^2 - 2 \times 7 \times 10 \times \cos B\)
A1.7 for the final answer of 15.3

1. **Determine the interior angle \(\angle PQR\):**
- Draw a North line at \(P\) and \(Q\).
- Since the bearing of \(Q\) from \(P\) is \(050^\circ\), the angle from the South line at \(Q\) to the line \(QP\) is \(050^\circ\) (alternate angles).
- This means the bearing of \(P\) from \(Q\) is \(180^\circ + 050^\circ = 230^\circ\).
- The bearing of \(R\) from \(Q\) is given as \(110^\circ\).
- The interior angle \(\angle PQR\) is the angle between the lines \(QP\) (bearing \(230^\circ\)) and \(QR\) (bearing \(110^\circ\)):
\[\angle PQR = 230^\circ - 110^\circ = 120^\circ\]

2. **Apply the Cosine Rule to find \(PR\):**
\[PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR)\]
\[PR^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(120^\circ)\]
We know \(120^2 = 14400\), \(150^2 = 22500\), and \(\cos(120^\circ) = -0.5\):
\[PR^2 = 14400 + 22500 - 36000 \times (-0.5)\]
\[PR^2 = 36900 + 18000 = 54900\]
\[PR = \sqrt{54900} \approx 234.307\text{ m}\]

3. **Round to the nearest meter:**
\[PR \approx 234\text{ m}\]

M1 for finding the correct interior angle \(\angle PQR = 120^\circ\)
M1 for substituting their values into the Cosine Rule: \(PR^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos(120^\circ)\)
M1 for evaluating \(PR^2 = 54900\)
A1.7 for the final answer of 234

1. **Use the Sine Rule to find the angle \(\angle XZY\):**
\[\frac{\sin(\angle XZY)}{XY} = \frac{\sin(\angle YXZ)}{YZ}\]
\[\frac{\sin(\angle XZY)}{8} = \frac{\sin(42^\circ)}{11}\]
\[\sin(\angle XZY) = \frac{8 \sin(42^\circ)}{11} \approx \frac{8 \times 0.66913}{11} \approx 0.48664\]
\[\angle XZY \approx \arcsin(0.48664) \approx 29.12^\circ\]
*(Note: Since the side opposite \(42^\circ\) is \(11\) cm, which is larger than \(8\) cm, the angle opposite \(8\) cm must be smaller than \(42^\circ\). Thus, only the acute angle \(29.12^\circ\) is geometrically valid, as \(180^\circ - 29.12^\circ = 150.88^\circ\) would lead to an angle sum exceeding \(180^\circ\).)*

2. **Find the angle \(\angle XYZ\):**
\[\angle XYZ = 180^\circ - 42^\circ - 29.12^\circ = 108.88^\circ\]

3. **Use the Sine Rule again to find \(XZ\):**
\[\frac{XZ}{\sin(\angle XYZ)} = \frac{YZ}{\sin(\angle YXZ)}\]
\[XZ = \frac{11 \times \sin(108.88^\circ)}{\sin(42^\circ)}\]
\[XZ \approx \frac{11 \times 0.94623}{0.66913} \approx 15.556\text{ cm}\]

4. **Round to 3 significant figures:**
\[XZ \approx 15.6\text{ cm}\]

M1 for applying the Sine Rule to find angle \(XZY\): \(\frac{\sin Z}{8} = \frac{\sin 42^\circ}{11}\)
M1 for calculating angle \(XZY \approx 29.1^\circ\) and hence angle \(XYZ \approx 108.9^\circ\)
M1 for applying the Sine Rule to find side \(XZ\): \(\frac{XZ}{\sin(108.9^\circ)} = \frac{11}{\sin 42^\circ}\)
A1.7 for the final length of 15.6